OxfordAQA A-Level Chemistry: Carboxylic acids and derivatives (9620)

This guide works through section 3.3.9 (Carboxylic acids and derivatives) of OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. Both subsections are included (3.3.9.1 on acids and esters, 3.3.9.2 on acylation), along with required practical 10. Because 3.3.9 is International A2 organic content, all of it is International A-level only. Its questions sit in Unit 4 (Organic 2 and Physical 2), which covers sections 3.3.7–3.3.16, while Unit 5 (Practical and synoptic) assesses all content.

Read this guide together with the revision notes and the practice questions. Background you need: making carboxylic acids by oxidation in alcohols, the carbonate test on the organic analysis page, and weak-acid behaviour on the acids and bases page. For the wider course, use the hub page, the printable checklist and the free diagnostics.

What this section covers

Section	What you must know or do	Stage
3.3.9.1	Structures of carboxylic acids and esters; acids are weak but liberate CO₂ from carbonates; esterification with an acid catalyst; uses of esters	International A-level only
3.3.9.1	Oils and fats as esters of propane-1,2,3-triol (glycerol); acid and alkaline hydrolysis; soap and glycerol; biodiesel from vegetable oils and methanol	International A-level only
3.3.9.2	Structures of acid anhydrides, acyl chlorides and amides; their reactions with water, alcohols, ammonia and primary amines; the addition–elimination mechanism for acyl chlorides	International A-level only
3.3.9.2	Industrial advantages of ethanoic anhydride over ethanoyl chloride in making aspirin	International A-level only
Required practical 10	Make a pure organic solid, then check how pure it is	International A-level only

3.3.9.1 Carboxylic acids and esters

Structures

A carboxylic acid has the –COOH group: a carbon carrying both a C=O and an O–H. That carbon is C1 of the chain, and the name ends in -oic acid. Butanoic acid is CH₃CH₂CH₂COOH.

An ester has the –COO– link between two carbon groups. Its name has two words: the alcohol part first (methyl, ethyl, propyl…), then the acid part ending -oate. The chain holding the C=O carbon is always the acid part.

Worked example 1: naming and splitting an ester

Name CH₃CH₂CH₂COOCH₂CH₂CH₃ and give the acid and alcohol that form it.

1. Find the C=O carbon. CH₃CH₂CH₂CO– has four carbons, so the acid part is butanoate.
2. The group on the single-bonded oxygen is –CH₂CH₂CH₃, so the alcohol part is propyl.
3. Name: propyl butanoate. It forms from butanoic acid and propan-1-ol.

Weak acids that still release CO₂

Only a small fraction of carboxylic acid molecules ionise in water, which makes them weak acids:

CH₃CH₂CH₂COOH ⇌ CH₃CH₂CH₂COO⁻ + H⁺

Even so, it releases CO₂ from carbonates: you see fizzing, and the gas turns limewater milky.

Worked example 2: methanoic acid and magnesium carbonate

Write the equation.

1. Methanoic acid is HCOOH, so the salt contains the methanoate ion, HCOO⁻.
2. Mg²⁺ needs two methanoate ions, so you need two acid molecules.

2HCOOH + MgCO₃ → (HCOO)₂Mg + CO₂ + H₂O

Making esters

Warming a carboxylic acid with an alcohol and an acid catalyst (usually a few drops of concentrated sulfuric acid) gives an ester and water. The reaction is reversible, so you write ⇌ and the yield stays below 100%.

Worked example 3: esterification yield

A student heats 5.28 g of butanoic acid with excess propan-1-ol and an acid catalyst, and collects 4.68 g of propyl butanoate. Calculate the percentage yield. (Ar: H 1.0, C 12.0, O 16.0)

1. Equation: CH₃CH₂CH₂COOH + CH₃CH₂CH₂OH ⇌ CH₃CH₂CH₂COOCH₂CH₂CH₃ + H₂O. The ratio is 1 : 1.
2. Mr of butanoic acid, C₄H₈O₂ = 88.0. Amount = 5.28 ÷ 88.0 = 0.0600 mol.
3. Mr of propyl butanoate, C₇H₁₄O₂ = 130.0. Maximum mass = 0.0600 × 130.0 = 7.80 g.
4. Yield = 4.68 ÷ 7.80 × 100 = 60.0%.

Uses of esters

The specification names the common uses: solvents, plasticisers, perfumes and food flavourings.

Oils and fats

Every vegetable oil and animal fat is an ester of propane-1,2,3-triol, also called glycerol, HOCH₂CH(OH)CH₂OH. Each of its three –OH groups is esterified by a long-chain carboxylic acid, giving three ester links per molecule.

Hydrolysis of esters

Hydrolysis splits the ester link, in acid or in alkali.

Conditions	Products from propyl butanoate	Extent
Heat under reflux with dilute acid	Butanoic acid + propan-1-ol	Reversible
Heat under reflux with NaOH(aq)	Sodium butanoate + propan-1-ol	Goes to completion

CH₃CH₂CH₂COOCH₂CH₂CH₃ + NaOH → CH₃CH₂CH₂COONa + CH₃CH₂CH₂OH

In alkali the acid ends up as its carboxylate ion, which does not react with the alcohol, so the ester cannot re-form. Add a strong acid afterwards to recover the free acid.

Soap, glycerol and biodiesel

Soap. When an oil or fat is hydrolysed by alkali, the products are glycerol and soap, made up of the salts of long-chain carboxylic acids.

Biodiesel consists of several methyl esters of long-chain carboxylic acids. To make it, a vegetable oil is reacted with methanol, with a catalyst present. Each acid chain becomes a methyl ester, and glycerol is released.

Worked example 4: one fat, two products

Glyceryl tritetradecanoate, (C₁₃H₂₇COO)₃C₃H₅, has Mr 722.0. Separate 28.88 g samples are (a) boiled with excess NaOH(aq) and (b) reacted with methanol and a catalyst. Calculate the maximum mass of sodium tetradecanoate (Mr 250.0), of methyl tetradecanoate (Mr 242.0) and of glycerol (Mr 92.0) from one sample.

1. Amount of fat = 28.88 ÷ 722.0 = 0.0400 mol.
2. (C₁₃H₂₇COO)₃C₃H₅ + 3NaOH → 3C₁₃H₂₇COONa + C₃H₅(OH)₃
3. Sodium tetradecanoate = 3 × 0.0400 = 0.120 mol, so mass = 0.120 × 250.0 = 30.0 g.
4. (C₁₃H₂₇COO)₃C₃H₅ + 3CH₃OH → 3C₁₃H₂₇COOCH₃ + C₃H₅(OH)₃
5. Methyl tetradecanoate = 0.120 mol, so mass = 0.120 × 242.0 = 29.0 g.
6. Glycerol = 0.0400 mol in either reaction, so mass = 0.0400 × 92.0 = 3.68 g.

3.3.9.2 Acylation

Structures

Family	Functional group	Example
Acyl chloride	–COCl	CH₃CH₂CH₂CH₂COCl, pentanoyl chloride
Acid anhydride	–CO–O–CO–	(CH₃CH₂CH₂CO)₂O, butanoic anhydride
Amide	–CONH₂, or –CONHR from a primary amine	CH₃CH₂CH₂CH₂CONH₂, pentanamide

An acyl group is R–CO–. Acylation replaces an H on O or N with an acyl group.

The four nucleophiles

Each of these two reagent families reacts with four nucleophiles: water, alcohols, ammonia and primary amines. The specification calls these nucleophilic addition–elimination reactions. An acyl chloride releases HCl; an acid anhydride releases a carboxylic acid.

Nucleophile	Pentanoyl chloride	Butanoic anhydride
Water	C₄H₉COCl + H₂O → C₄H₉COOH + HCl	(C₃H₇CO)₂O + H₂O → 2C₃H₇COOH
Propan-1-ol	C₄H₉COCl + C₃H₇OH → C₄H₉COOC₃H₇ + HCl	(C₃H₇CO)₂O + C₃H₇OH → C₃H₇COOC₃H₇ + C₃H₇COOH
Ammonia	C₄H₉COCl + 2NH₃ → C₄H₉CONH₂ + NH₄Cl	(C₃H₇CO)₂O + 2NH₃ → C₃H₇CONH₂ + C₃H₇COONH₄
Ethylamine	C₄H₉COCl + 2C₂H₅NH₂ → C₄H₉CONHC₂H₅ + C₂H₅NH₃Cl	(C₃H₇CO)₂O + 2C₂H₅NH₂ → C₃H₇CONHC₂H₅ + C₃H₇COO⁻ ⁺H₃NC₂H₅

C₄H₉ and C₃H₇ are straight chains. The anhydride route makes propyl butanoate without the equilibrium that limited Worked example 3. With ammonia or an amine, a second molecule acts as a base and takes up the acid by-product.

Mechanism: nucleophilic addition–elimination

You must outline this mechanism for acyl chlorides with all four nucleophiles. The carbonyl carbon is strongly δ+ because O and Cl both draw electrons from it. The nucleophile attacks with a lone pair on O (water, alcohols) or N (ammonia, amines).

Worked example 5: pentanoyl chloride and ethylamine

Addition
  arrow 1: lone pair on N of CH3CH2NH2  →  carbonyl C (δ+)
  arrow 2: C=O π bond                   →  O, which becomes O⁻

  intermediate:  C4H9–C(O⁻)(Cl)–N⁺H2CH2CH3

Elimination
  arrow 3: lone pair on O⁻  →  re-forms C=O
  arrow 4: C–Cl bond        →  Cl, which leaves as Cl⁻
  arrow 5: a second CH3CH2NH2 removes H⁺ from N⁺

  products:  C4H9CONHCH2CH3 (N-ethylpentanamide) + CH3CH2NH3⁺Cl⁻

Each arrow begins at a bond or a lone pair. With water or an alcohol the steps are the same, but the positive charge sits on oxygen and Cl⁻ takes the H⁺ to give HCl.

Aspirin: why industry uses ethanoic anhydride

Aspirin is made by acylating the –OH group of 2-hydroxybenzoic acid:

HOC₆H₄COOH + (CH₃CO)₂O → CH₃COOC₆H₄COOH + CH₃COOH

Ethanoic anhydride is preferred to ethanoyl chloride because it:

• is cheaper
• is less corrosive
• is less readily hydrolysed, so it copes better with water and damp air, making storage and handling easier
• reacts less vigorously, so the process is easier to control
• produces ethanoic acid, not corrosive, toxic hydrogen chloride gas, as its by-product.

Required practical 10: a pure organic solid

The specification lists “Prepare a pure organic solid and test its purity” as required practical 10, and states that questions may be set on the practical or on the skills in it. Aspirin, made by the reaction above, is one solid you could prepare.

Purifying by recrystallisation

1. Filter off the crude solid under reduced pressure.
2. Dissolve it in the minimum volume of hot solvent, so the solution is saturated and most product crystallises on cooling.
3. Filter the hot solution to remove any insoluble impurities.
4. Cool the filtrate. Crystals form; the small amounts of soluble impurities remain dissolved.
5. Collect the crystals by filtration under reduced pressure and wash with a little cold solvent, which rinses off impure solution without dissolving much product.
6. Dry the crystals.

Testing purity

Compare its melting point with the data-book value. A pure solid melts sharply, over a narrow range, at the data-book value. Impurities lower the melting point and make it melt over a wider range.

Worked example 6: yield after purification

A student acylates 4.14 g of 2-hydroxybenzoic acid (Mr 138.0) with excess ethanoic anhydride. After recrystallisation, 4.05 g of dry aspirin (Mr 180.0) is obtained. Calculate the percentage yield.

1. Amount of 2-hydroxybenzoic acid = 4.14 ÷ 138.0 = 0.0300 mol.
2. The ratio is 1 : 1, so maximum mass of aspirin = 0.0300 × 180.0 = 5.40 g.
3. Yield = 4.05 ÷ 5.40 × 100 = 75.0%.

Some product always stays dissolved in the cold solvent.

Common errors

• Naming an ester backwards. The alcohol part comes first: CH₃COOCH₂CH₂CH₂CH₃ is butyl ethanoate, not ethyl butanoate.
• Calling a carboxylic acid strong because it fizzes with a carbonate. It is weak because it dissociates only partly.
• Giving the free acid as the product of alkaline hydrolysis. You get its salt.
• Treating fat and soap (or methyl ester) as 1 : 1. Each fat molecule gives three.
• Writing HCl as a product when an acid anhydride reacts. The by-product is a carboxylic acid.
• Using one NH₃ or one amine molecule per acyl chloride. You need two.
• Calling the mechanism nucleophilic substitution, or leaving the charges off the intermediate.

Where to go next

Condense this section with the revision notes, then test yourself with the practice questions and the free diagnostics.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.9 Carboxylic acids and derivatives (3.3.9.1 and 3.3.9.2) and required practical 10, International A2 content.